Question

An infinite plane of charge has a surface charge density of 5 µC/m2 . How far apart are the equipotential surfaces whose potentials differ by 105 V? The permittivity of free space is 8.85 × 10−12 C 2 /N · m2 . Answer in units of mm.

Answer 1

Distance in mm will be 0.3718 mm

Step-by-step explanation:

We have given charge surface charge density
`\rho _s=5\mu c/m^2=5* 10^(-6)\mu c/m^2`

We know that electric field due to surface charge density is given by

`E=(\rho _S)/(2\epsilon _0)=(5* 10^(-6))/(2* 8.85* 10^(-12))=2.824* 10^5Volt/m`

We have given potential difference V = 105 volt

We know that potential difference is given by
`V=Ed`

So
`105=2.824* 10^5* d`

`d=37.181* 10^(-5)m=0.3718mm`